Problem in understanding the process of calculating the rotational inertia As we know, rotational inertia is the mass-equivalent in rotation.  

For a discrete body, it is measured as $$I = \sum m_i{r_i}^2 $$ . But when a continuous body comes, $$I = \int r^2 .dm$$ which implies $$I = \phi + C$$ where $\phi$ is the function that gives the change and C is the initial point. 

This is what my book writes. Ok, that's the pure definition of indefinite integral. But at what I am confused is that what is $\phi$ dependent on and what is $C$ here. $C$ , in general, gives us the initial condition so as to give us the original function. But what is $C$ representing here?? What is $\phi$ dependent on?? 
If we use proper limits , then it becomes $$ I = \int_{a}^{b} r^2.dm \implies I = [\phi(b) + C] - [\phi(a) + C]$$ . It is different that $C$ is cancelled here. But it does exist. So what does this $C$ represent in case of calculation of moment of inertia??
Please explain. 
 A: Unless I have the rest of the book, it is difficult to understand what he meant. But many books have equations that could make not much sense or whose explanations are very unclear (no author is perfect). The only thing you really need to know is what you already seem to know. In an actual calculation you use a definite integral, and the constant $C$ doesn't matter, it always goes away.
My guess is that the author meant that $C = \phi(a)$, because you can write
$$I = \int_{a}^{x} r^2.dm=\phi(x) - \phi(a)$$
A: I think the book authors might have the Huygens-Steiner theorem in mind when saying that.
The theorem says that if the body has moment of inertia $I_{cm}$ with respect to axis crossing its center of mass, then moving the reference axis to another parallel axis gives a new moment of inertia $I'$, related to original one by
$$I'=I_{cm}+md^2,$$
where $d$ is distance between axes. So I suppose the $C$ in your quotation is $md^2$.
I wouldn't think that it really is the constant of integration, because normally moment of inertia is not an indefinite integral, and usually not even a definite one: it's instead a multiple integral.
A: Where are you getting these books? Here it makes absolutely no sense to mention indefinite integrals and I have never seen such confusing statements.
You need definite integral, usually in multiple dimensions (ideally 3D, as we live in 3D space, but you can simplify in some cases). The integration goes over the entire volume of the object, you have $dm=\rho\,dV$ and if the density is constant, it's just a regular integral over volume. No additional constants, nothing. Indefinite integral in multiple dimensions is a weird concept anyway.
However, result depends on selection of your rotational axis (usually you make it go through coordinate origin so $r$ can be the $\sqrt{x^2+y^2}$ if you are looking for the moment of inertia around axis in the direction $z$). If you calculate the momentum of inertia around the center of mass, then you get the  inertia around another axis by using Steiner's law:
$$J=J^* + mr^{*2}$$
if $r^\ast$ is the distance between the axis and the object's center and $J^*$ is the moment of inertia around the mass center. However, this relation has nothing to do with indefinite integrals. It follows immediately from
$$\int(r+r^*)^2dm=\underbrace{\int r^2 dm}_{J^*}+2r^* \underbrace{\int r dm}_{\text{mass center}=0}+r^{*2}\underbrace{\int dm}_m$$
You see that you only get Steiner law if your $r$ coordinate frame is put in the object's center (otherwise the middle term would not be 0).
So... find another book.
